Response of First Order RL Circuits Part 1-Electric Circuits 8th Edition Nilsson-Electrical Circuital Analysis-Solution Manual, Exercises of Electronic Circuits Analysis

Download Response of First Order RL Circuits Part 1-Electric Circuits 8th Edition Nilsson-Electrical Circuital Analysis-Solution Manual and more Exercises Electronic Circuits Analysis in PDF only on Docsity! 7 Response of First-Order RI and RC Circuits Assessment Problems AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit. AWA WMA 30 60 1z0v@ v 2300 T - i(07) First combine the 302 and 6 resistors in parallel: 30|6 = 52 Use voliage division to find the voltage drop across the parallel resistors: v= 543 (120) =75V Now find the current using Ohm’s law: (0) = 2 = BO i(0-) = aia. [b] w(0) = 5L%(0) = 58 x 107)(12.5)? = 625 mJ [c] To find the time constant, we need to find the equivalent resistance seen by the inductor for t > 0. When the switch opens, only the 2 resistor remains connected to the inductor. Thus, ab 8x10, “RO 2 mS [d] a(t) = i(0-)el/7 = —12.5e71/9 — 12. 5e°- 2" At > 0 [e] #(5 ms) = —12.5e7?50(0005) — _19 5e-125 — -3.58A 7-1 docsity.com 7-2 CHAPTER 7. Response of First-Order RL and RC Circuits So w(5ms) = $Li?(5ms) = }(8) x 10-9(3.58)? = 51.3mJ w (dis) = 625 — 51.3 = 573.7mJ % dissipated = () 100 = 91.8% AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor: + 6.4A(8 ve 100 J Using current division, 10 iO") = Hyg) =4A Now use the circuit for t > 0 to find the equivalent resistance seen by the inductor, and use this value to find the time constant: AW 6f) + we 109 J fo-328 240 i Req =4||(6+10)=320, «. r=e = Use the initial inductor current and the time constant to find the current in the inductor: a(t) = i(0-)e*/7 = 4e#/91 — ge“ At > 0 Use current division to find the current in the 10 resistor: 4 i (t) = —1 tlt) = T5046 Finally, use Ohm’s law to find the voltage drop across the 10 resistor: Vo(t) = 10%, = 10(—0.8e71%) = —8e%V, t> 0+ [b] The initial energy stored in the inductor is =0.1s . 4 —100 —10t + (-i) = 39 (4 y=-08e A, t>0 w(0) = 5Li2(0") = 5 (0-32)(4)? =2.56J Find the energy dissipated in the 4 resistor by integrating the power over all time: v4o(t) = Le = 0.32(—10)(4e7™) = —12.8e7-1* v, t>0* 3 docsity.com Problems = 7-5 i(0-) = 24/2 = 12.A = i(0*) Note that i(0~) = i(0*) because the current in an inductor is continuous. [b] Use the circuit at t = 0+, shown below, to calculate the voltage drop across the inductor at 0*. Note that this is the same as the voltage drop across the 10 resistor, which has current from two sources — 8 A from the current source and 12 A from the initial current through the inductor. * vor) $102 Wea aaal| — 200mH| v(0+) = —10(8 + 12) = —200V {c] To calculate the time constant we need the equivalent resistance seen by the inductor for t > 0. Only the 10Q resistor is connected to the inductor for t > 0. Thus, 7 = L/R = (200 x 1077/10) = 20ms {d] To find i(t), we need to find the final value of the current in the inductor. When the switch has been in position a for a long time, the circuit reduces to the one below: ie} 2100 Waa Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit. Thus, ip =—8A Now, a(t) = is + [i(0+) —isle/” = —8 + [12 — (-8)]e"/0- =—-8+20e5* A, t>0 [e] To find v(t), use the relationship between voltage and current for an inductor: v(t) = = = (200 x 107*)(—50)(20e°") = —200e°* V, st > OF 3 docsity.com 7-6 CHAPTER 7. Response of First-Order RL and RC Circuits Wie AWA AP 7.6 [al + 8kN + 40kQ 0.25uF zy v7 160k DIV ° v From Example 7.6, Vo(t) = —60 + 902710 V Write a KVL equation at the top node and use it to find the relationship between v, and v4: VA — Vo VA vat 8000 160,000 * 40,000 — 20u4 — 20u5 + v4 + 4u,4 + 300 = 0 25v4 = 20v, — 300 va = 0.8v, — 12 Use the above equation for v, in terms of v, to find the expression for v4: va(t) = 0.8(—60 + 90e71) — 12 = -60+ 72e7"V, *t>0t [b] t > Ot, since there is no requirement that the voltage be continuous in a resistor. AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop across the capacitor: 60kQ AW iL + 1oma) 40kQS v{0-) 225kQ ._ (40 x 108 ~ \125 x 108 ue(0~) = (3.2 x 107*)(25 x 10°) = 80V so v,(0+) = 80V Now use the next circuit, valid for 0 < t < 10 ms, to calculate v,(t) for ) (10 x 10-3) = 3.2mA that interval: 60kQ e ANN . 25kQz WE Tv, e 3 docsity.com Problems = 7-7 For 0<t<100ms: 7 = RC = (25 x 10°)(1 x 10~°) = 25 ms v(t) = ve(0-)e/" = 8004" V 0<t < 10ms [b] Calculate the starting capacitor voltage in the interval t > 10 ms, using the capacitor voltage from the previous interval: v¢(0.01) = 80e~40-!) — 53,63 V Now use the next circuit, valid for ¢ > 10 ms, to calculate v,(t) for that interval: a 25kQ 53 .63V ¥ weTv, #100kQ c For t>10ms: Req = 25kQ||100kQ = 20kQ T = RegC = (20 x 10%)(1 x 107%) = 0.028 Therefore v,(t) = ve(0.01* )e~¢-090/" = 53.63e- 5-9") Vt > 0.018 [c] To calculate the energy dissipated in the 25 kQ resistor, integrate the power absorbed by the resistor over all time. Use the expression p= v"/R to calculate the power absorbed by the resistor. 0.01 (gde~4e]2 co [53,63 50t-0.01)]2 Mask = I 25,000 4° + Do 25,000 [d] Repeat the process in part (c), but recognize that the voltage across this resistor is non-zero only for the second interval: 22 [53.63e~t-0.01)]2 0.01 100,000 ‘We can check our answers by calculating the initial energy stored in the capacitor. All of this energy must eventually be dissipated by the 25kQ resistor and the 100 kQ resistor. Check: Wstorea = (1/2)(1 x 107®)(80)? = 3.2mJ dt = 2.91mJ Woke = dt = 0.29mJ Weiss = 2.91 + 0.29 = 3.2mJ docsity.com 7-10 CHAPTER 7. Response of First-Order RL and RC Circuits Up = —20(5)(t — 32 x 1073) + 8 = —100¢ + 11.2 The output will saturate at the negative power supply value: —15=—100¢4+ 11.2 -. ¢ = 262ms AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: Up = V7 + [Vo — Vsle/” = —2 + (0 + 2)e-*/” 7 = (160 x 10°)(10 x 10°) = 107°; 1/r = 625 Up = —2+ Qe Ot V7. Un = Up Write a KVL equation at the inverting input, and use it to determine v,: Un. Un — Vo 70,000 * 40,000 ~ Vo = 5Un = 5Up = —10 + 10e~85' v The output will saturate at the negative power supply value: —10 + 10e~* = —5; ef = 1/2; ¢=In2/625 = 1.11ms [b] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: Up = Vz + [Vo — Velo’ = —2 + (1 + 2)e-O = —2 + 3c V The analysis for v, is the same as in part (a): Vo = Bp = —10 + 15e7 7! V The output will saturate at the negative power supply value: 10+ 15" = 5; eo 1 1/3; t =n 3/625 = 1.76 ms docsity.com Problems 7-11 Problems P71 ~~ [aj ¢(0) = 125/25=5A L 4 =—=-——=40ms [b] + R= i00 Oms [c] i=5e*#A, = t>0 vy = —80i = —400e* Vt > 0 V2 = pet 4(—125e77") = —500e" Vs t > OF dt [d] Paiss = ?(20) = 25e75%(20) = 500e-5" W ee ft 800 a eg ne | — 19 e—50t dias =-[ 500e"% dar = 500 | = 10 — 1068 J Waiss(12 ms) = 10 — 10e~°* = 4.515 w(0) = 5()(25) =50J 451 % dissipated = <* (100) = 9.02% P72 [alt<o 15k 15k Wi AW > igo) 2 igo) ve J? 18k igo ) 15kQI||15kQ = 7.5k2 fie 9 (0) = Geyray xi ~o4mA 4,(0-) = %2(0-) = (0.4) x 10 =0.2mA [b] 41(0*) = %(0-) = 0.2mA i2(0*) = —i,(0*) =-0.2mA —_ (when switch is open) L _ 30x10 R~ 30x 10 a(t) = i(0*)e“”” ix(t)=0.2e2 mA, t>0 3 docsity.com [e] r= =10%, 2-108 T 7-12 CHAPTER 7. Response of First-Order RL and RC Circuits [d] to(t) =—ii(t) when ¢>0* +. g(t) =—0.2e°°%tmA, t>0* {e] The current in a resistor can change instantaneously. The switching operation forces i2(0~) to equal 0.2mA and i2(0+) = —0.2mA. P73 [a] i.(07)=0 __ since the switch is open for t < 0. [b] For t = 0~ the circuit is: : iy Wie WA 4 ANA 500 4000 2000 25v@) 23000 - Ok 0") ; - 300 2||3002 = 1502 5 25 to = 504150 = 125mA - dae! 300) . i,(0-) = (Fan) tg = 62.5mA [c] For t = 0+ the circuit is: =i lg AW ANA, Wi 502 1002 2002 + 25vC = 3000 v(0*) Q62.5ma igo+) 300 Q|1002 = 752 25 fo = spy 7g = 200mA 7300 in = (=) 200 = 150mA i,(0*) = 150 — 62.5 = 87.5mA [d] i(0*) = i,(0-) = 62.5mA [e] t.(00) = tg = 150mA docsity.com Problems fb] in = 22°; r= z = at x 10-3 = 1ms ty, = 2e- 100% A ig =4-ip =4—2e MA, = t > 0 [c] 4 — 2e7 100 — 3.8 0.2 = Qe 1000 1000 _ 10 “. ¢=2.30ms P78 [a] For t<0 15 50 30 We WA AM >} + g Yo z = sove [3 00 J 760 — F200 ijo) igo} 20 ANN i= = =2A -, _ 2(50 . i,(0-) = ion =1A =i,(0*) For t>0 AWN 3Q = 2 602 F200 omg. 20 AWN iz(t)=iz(Ot)e“7 A, t>0 L 020 1 = R- 54157 10078 iz(0t) =1A in(t}=e A, t>0 vo(t) = —15%,(t) Vo(t) = —15e71 vy, t>0* 7-15 docsity.com 7-16 CHAPTER 7. Response of First-Order RL and RC Circuits 2 P79 Pro = a = 11.2562 0.01 Wein = i 126% dt _ 11.25 200 (" ~ —200 0 = 56.25 x 10-3(1 — e~?) = 48.64 mJ Wstored = 502»? = 100mJ. 48.64 % diss = = 59- x 100 = 48.64% P71 t 7.10 [a] t<0 Me 2 -5mAC = 16k0 igo) * _, _ —2.5(16) i,(0-) = (0) 2mA t>0 + + ar v2 1k0 omH > omit | v, 2 1k0 -2mA ~2mA _ — 0x10" _ 49% 10°; 1/7 = 25,000 rT T98 i Te) Vo = —1000(—2 x 10~*)e~ 75,0008 —. 2¢-25,0001 yy t>or [b] wae = +(40 x 1079)(4 x 10~°) = 80nJ [c] 0.95wae = 76 nd to 4e@—50,000t 76x 10° = [ wn 0 docsity.com Problems 76 x 10-° = 80 x 10-%e0 | = 80 x 10-9(1 — e000) 90.0005" 0.05 50,000, =In20 so t,=59.9ps t. 59.9 _ ; = 7a 7 1.498 so t,% 1.57 P71l t<0: 7 50 200 ANA, Ar WA 10A 160 $50 2A in(0t) =2A t>0: 52 200 NA Wie a vz 150 250 osm -_ i, i, (20)(5) Re = Ga + 20 = 242 L 96 lig 1 T= ay XO =4ms; = = 250 iy = 2e7 7 A ep Ole —250t ) tom ope = 0.4e A Up = —15ip = —6e™%* VY, *¢>0t P 7.12 poon = 2012, = 20(4)(e~>™)? = 80e2" W Siee e500 oo wooo = f 8062 gt = 80° __ op |, = 160mJ w(0) = 5 (96)(10-*)(4) = 192mJ i % diss = jp (100) = 83.33% TAT docsity.com 7-20 CHAPTER 7. Response of First-Order RL and RC Circuits t>0: Din Kote + \z100n = 2000 Vy ix in = in (200) _ 2 a 300” 3°" ;,(100)(200) _ 59, 2 | 200, = 50ia + 300 50trs + 3 °T BF Rg 4 = 1000 ip 3 3 + a, ¥, f 200mH = 1002 L200 1 == x 1073 os T= = Tog 0 ; 500 ip =6e "A, tt > 0 [b] vz = 200 x 1073(—3000e7°) = —600e >" V, t > oF le] S0i, Fe - “, 2 1000 2000 i & ut = 50ig + 100i, = 150%, i UL _ _ 46-5008 > ot a= 50 = Ae A t>0 P7.16 w(0)= 5 (200 x 10~*)(36) = 3.63 Psoig = —B0igiz, = —50(—4e7) (GeO) — 1200e7 10% W e7 10008 °° iim = f 1200¢~ 1° dt = 120055 |, = 1.25 % dissipated = $2 (100) = 33.33% docsity.com Problems 7-21 P717 [al t<0 igo) 2. e- 120 S40kQ $20kKQ F60kQ $30kQ 40 kQ||20kQ = 13.33k0 60kQ|]30kQ = 20k (120 x 107%) (13.33 x 10%) = 1600 V igo) — 13.33kQ 1600VC 20kQ ope 1600 i,(0)= 33,333.33 =48mA ae 200mH — > 48ma 20kN= = 60k L 0.2 1 3a 2.5 ys; = = 400,000 iz (t) = 48e7400000 mA, t>0 Poor = (0.048e~4°%-)? (60,000) = 138.24e~ 890.0% Wy t Weiss = f 138.24e~800.000 doy — 172.8 x 10-8[1 — @~ 800.0004) J 0 w(0) = 5(2)(48 x 1079)? = 230.4 pI 0.25w(0) = 57.6 pd 172.8(1 — e7 800.000") — 57.6; £5 OO 1S Inl5 = 300,000 = 2307 Hs docsity.com 7-22 CHAPTER 7. Response of First-Order RL and RC Circuits [b] waiss(total) = 230.4(1 — 7800000) 1,5 Waiss(0.507 pus) = 76.82 J % = (76.82/230.4) (100) = 33.38% e 10 P7.18 [a] t<0: vc — — 20a 220vC (50/3)A212N (10/3)Az600 o =ot: 10 t=0': Mw—~€ — 220A 220vE Ne (50/3) aq (10/3) aq re 220 = ta» + (50/3) + (10/3), ia» = 200A, t=0* [b] At t=oo: a a — 20a 220vC R $120 2602 & ov tap = 220/1 = 220A, t=00 10 Wie — 220A vow ym Li 1 ae 15mH {c] 4:(0) =50/3, n= i x 10-$ = 0.167 ms #2(0)=10/3, m= a x 10-3 = 0.25ms ix(t) = (50/3)e- "A, t >0 docsity.com Problems 7-25 i= x 108 = 5c ma, tot 5 5kQ + — >i + 2uUF 7 vy v, 7 Suk —108 -3 91252 —125¢ n= 5 f 19x10 5 de + 75 = BOE" 415V, t>0 _ 10° —3 ,—-1252 —125t 0 a [15 x10" e125 dp + 0 = be" 415V, t>0 [b] w(0) = 32 x 107%) (5625) = 5625 J [c] wrapped = 52 x 10~®)(225) + 58 x 107%)225 = 1125 pJ. Weiss = 5(16 x 107°) (5625) = 4500 pJ. Check: Wtrapped + Waiss = 1125 + 4500 = 5625 pJ; w(0) = 5625 pJ. P7.22 [a] R= 7 = 20kQ L L DB] > = aq = 10005 C= Tapa x 108) ~ O05 HF 1 [ce] 7 = 799 = 1 ms {d] w(0) = 5 (0.05 x 1076)(10*) = 250 pJ le] to vw to (108) 7 Waiss = f, at = f “(20 x 103) Ss to _ - — 62000 = 0.5559 i 250(1 —e ) ps 200 = 250(1 — e7 200%) 67200 _ 9.9. 62000to _ 5 to= £ Ind; to = 804.72 pus °= 5009 ni ‘0 © 804.72) 3 docsity.com 7-26 P 7.23 P 7.24 CHAPTER 7. Response of First-Order RL and RC Circuits [a] <0: 2v 20kQ 5kQ 10kQ O Me We AWN ig0-) — ig0-) 10v@) =5kQ 8 og i,(0-) = i2(0-) = (a x10 ) =0.2mA [b] t > 0: 0. 4uF ——it— + 2V - 5kQ | 10kQ Wy WW ijo-) if0-) 4 : 2 5kQ 2 (Ot) = oi 44(0 )= 79 x 10 0.2mA ig(0*) = = 10-8 = -0.2mA 10 {c] Capacitor voltage cannot change instantaneously, therefore, 4(0-) =%4,(0t) =0.2mA [d] Switching can cause an instantaneous change in the current in a resistive branch. In this circuit i2(0-) =0.2mA and i2(0*) =—0.2mA [e] % =2e/7V, t>0 7 = RC = 5000(0.4) x 10-° = 2 x 10° Uc = 2eFO Vy, t>0 1= ae = 0.2e5 mA, t>0 f] @=s5 a =—0.2emA, *t>0+ [a] v(0) = as =118.80V Re= es) = 2k docsity.com Problems 1 T = R.C = (2000)(0.25) x 10-* = 500 ps; = = 2000 v=118.80e7™ Vt >0 Vv _ = —2000t = 3000 39.6e mA to [b] w(0) = 5 (0-25)(118.80)? = 1764.18 pd —2000¢ ae = SOC = 19.82 mA 6 Par = [(19.8)e~2°}? (4000) x 10-6 = 1568.16 x 1078740 740002 )250x10-6 Wa = 1568.16 x 10-° = 392.04(1 — e!) pJ —4000 lo 392.04 1 — _ 100 = 14. % = Tagg (t ~ ©) X 100 = 14.05% P7.25 [a] t<0: kN AAA ar + + 25maQ) vf 3 8k v {07 8 10kQ i oy (25)(8) _ i,(0-) = (20) =10mA v(0~) = (10)(10) = 100 V ig(0-) = 25-10 =15mA v2(07) = 15(8) = 120V t>0 2kQ L + + 120V 7 0.15uF | 0.30ur 7 = RC =0.2ms = 200 ps; 2 = 5000 7-27 docsity.com 7-30 CHAPTER 7. Response of First-Order RL and RC Circuits P7.27 t<0 ig 2x10 i, 3.3k0 o + 25v) v.07) =25V t>0 60kN vp = 2x 10*i, + 60,000i7 20,000(—ér) + 60,000é7 = 40,000é7 Il OF Roy = 40k uw 40kN = v 25v 7 25nF ° 7 = RC =1ms; = = 1000 Up = We 10008 y, t>0 d ip = 25 x 19°? 5, 251") = —625e71 WA, = t > OF docsity.com Problems P 7.28 [a] T= RC = Ry (0.2)x 10° =10°; Rew = pd =5kO 29KQ : (i, ove) Sl 2 1* * avy + vy vq 2 10k0 vp = 20 x 10°(ip — ava) + 10 x 10%p ug = 10 x 10%ip vr = 30 x 10%ip — 20 x 10%a10 x 10%p UT 7, = 30 x 10° — 200 x 10°a = 5 x 10° T 30 — 200,000a = 5; a= 125 x 107° A/V [b] v(0) = (0.018)(5000) =90V t<0 t>0: + + 9OV70.2uF vz5kN Uo = Me IHY, *+>0 ago eS + avy + 0.2uF Tv, vq = 10kQ VA VA — Vo vA, YA Vo Gy, = Tox ie + 3p.000 ~ 125 x 10Pva =0 Qua + va — Vo — 2500 x 10-v, = 0 “UA = 29 = 180e7 1 V 7-31 docsity.com 7-32 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.29 [al 20k -1000t 22.5e & + _ one ey ns 0.2ue F906 Vv tate v3 10k Pas = (—90e71) (22.5 x 10~%¢7 100) — —2025 x 1073¢~2000 yy iy [Pa dt = —1012.5 uJ. *. dependent source is delivering 1012.5 pJ __ (180)?e~200% [b] Pick = 10 x 103 wror = |” prow dt = 1620 uJ 0 _ (90)?e~200% Pook = "90 x 108 a toon = Poor dt = 202.5 wd 0 we(0) = 3002) x 107-6(90)? = 810 uJ SY waey = 810 + 1012.5 = 1822.5 pJ DY waies = 202.5 + 1620 = 1822.5 J. P 7.30 [a] At t= 0° the voltage on each capacitor will be 25 x 107 x 200 = 5 V, positive at the upper terminal. Hence at t > 0+ we have 1 + 4 #50 SV 7 S0UF 25maA(4) = 2000 - sd isa(Ot) = 0.025 + : + 7 = 165A docsity.com Problems 7-35 P 7.33 After making a Thévenin equivalent we have t=O ~ ig) ANA AMY 15kQ 5kQ ip 180v@) vgt) ¢250mH I, = 180/15 = 12mA 7 = (0.25/20) x 10-3 = 0.125 x 1074; = 80,000 Vv, _ 180 _ f= 3 = 20 =9mA to = 9 + (12 — 9)e~ 80,000 — 9 + 3e~ 80000 mA Vo = [180 — 12(20]}e~ 80 = —G0e~80.00% y P7.34 t<0 io-) 382 @ea i,(0-) =6A t>0 12Q 5mH 8Q AW i + AW - vy, + 32vG@ v, @aev , 32+ 48 iz(co) = 20 4A 3 docsity.com 7-36 P 7.35 CHAPTER 7. Response of First-Order RL and RC Circuits EB 5xi0% 1 T= Bp 59 = 250 pis; 7 = 4000 in =44+ (6 — 4)e 1% = 44 2e- AA => 0 Vo = —8iz + 48 = —8(4 + Ze 4) + 48 = 16—16e™ V, st > OF vy, = 5 x 10 = 5 x 10~3[—8000e-4'] = —40e~4" V, st > OF Check: at ¢ = 0* the circuit is: 12Q 6A 8Q v,(0+) =32—72+0=-40V, (0+) = 48-48 =0V fa] <0 5Q MA + 40aQ 40E vl0-)F 200 igo-) KVL equation at the top node: _ vo(07) , vo(07) , vo(07) —40 = Z Ge 20 + 5 Multiply by 20 and solve: —800 = (5+1+44)u,; Vo = —80V io(0-) = = = —80/5 = —16A docsity.com Problems 7-37 ue o 609 5Q Wy AW. a 240vC) vz 202 310m 7 4 o Use voltage division to find the Thévenin voltage: _ 20 ~ 20+ 60 Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance: Ryn = 5 + 20|/60 = 5+ 15 = 202 The simplified circuit is: Vin = Uo (240) = 60V 20Q AW 60ve 3 toms i} EL 10x 107% L Te 0.5 ms; a 2000 ‘ 60 tp(0o) = The 3A tg = i9(00) + [io(0*) — io(00)]e~/7 = 3+ (—16 — 3)e~2% — 3 19¢-2 At > 0 [Bl % = big + (0.01) 2 = 15 — 95e~700) + 0.01(38,000) (e720) = 15 — 95e~ 200 + 38Qe~2000" Vo = 15+ 285e72000 Vy t>or P 7.36 [a] For t <0, calculate the Thévenin equivalent for the circuit to the left and right of the 75 mH inductor. We get docsity.com 7-40 P 7.40 P 7.41 CHAPTER 7. Response of First-Order RL and RC Circuits v fal ¥@ sR f 4 -S+tecf- vdt+I=0 Differentiating both sides, ldv 1 Ra’z~° dv R aE ot: Tr =0 dv R {[b] TE” dv R R at = pat so du= yu dv R ee MO) da fee a ale vt) oR ee v(t) = Voe P/E — (V,-R. Iy)e~ P/E) Fort <0 se) sa + NGF 2002 v,G) S250 0.9 - ‘> Uz Uz — 250 VU, — 250] _ 5t9| athe 2 Ue, 4, (U2 — 250) a = + 10 50 =0 5uz — 5000 + 20v, = 0; Uz = 200V docsity.com Problems 7-41 io(0-) = 200/20 = 10A t>0 87.2mH 102 40Q Whe ahaa te * 2102 @z250v 20Q¢ 0.9¥5 O37 .5V 87.2mH 102 8Q AWN Me HG 200 ©sov 0 “Vy Find Thévenin equivalent with respect to a, b 10Q 8Q ae AW AM + Vpn 2)80v 0.9uy be Vin — 80 | , (Vin — 80) gith— 7-0 Vn=80V ist 18 T iy 10Q 8Q ae Nv AM + + Vg- vy wy 0.9vy be op = (ip —0.904)18 = [ix 0.9 (2)] 18 3 docsity.com 7-42 CHAPTER 7. Response of First-Order RL and RC Circuits vp =18ip —Yup Ov = 187 oF = Rey = 1.89 ir 87.2mH 1.8Q 2 We ig a 20Q¢ @eov & i9(00) = 80/21.8 = 3.67 A T= ere x 10-3 = 4ms; 1/7 = 250 21.8 ig = 3.67 + (10 — 3.67)! = 3.67 + 6.38e 7" A, t > 0 P742 t<0; 102 150 AM “Wii 2maf) =5Q ig) i,(0-) = —2---(0.002) = 0.5 mA ow BABS , ip(0*) =i,(0-) = 0.5mA t>0; 102 150 de AW AW i: * din 2mach) ¥, $50 vjo+)Qosma 240 <S Va — Vo Va - —0.002 + = + 15 =0 Vo — Va 4, Yo ge Zs 15 +5 x 10 +7 4i, — 0.001 = 0 docsity.com Problems 7-45 P 7.45 [al weiss = sLei*(0) = >(4)(25) =50J [b] 4 2 / 1 oom hon = 75 | «(40 202 dap +5 —100 6-20 |# 5 oe , 10 = ioe op pf) = 9° +aA t ion = ; [ (—400)e-2 de + 0 —200 e202 10 og 10 Gea = et _ A 3 20 |, 1° 3° 3 éapped = 5(18)(100/9) = 1003 [e] w(0) = 5(12)(25) = 1503 P746 [a] t<0 20aG) 15Q2 12H 1H 8 12 t>0 ro + fon ve48Q 10a 20a - re . 48 1 iz(0-) =i,(0*) = 20A; T= Gg = Os; 770 iz(oo) =10A ir = 10+ [20— 10Je?™ = 10+ 10e7!™ A, t>0 Vo = 4.8[—100e7 1] = —480e"" V, +¢>0t t [b] i: = af 48007!" dr + 8=4e7 44A, 420 t [e] io = if —A80e7? dr+12 =6e™46A, t>0 0 3 docsity.com 7-46 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.47 Fort <0, i4omu(0) = 75/5 = 15A For t > 0, after making a Thévenin equivalent we have : Wh ic— 40 & Vv °£ A0mH usa °) 100 v 60 mH V, 100 | L=15A; = =25A 4 = 25+ (15 —25)e™ =25-10e*" A, t 30 Up = oot = 0.04(400e-) = 16e-#"° V, = t > OF P 7.48 [a] 2(0-) = 5520) =24V C =(4+4) =100r ea \30 " 15) For t > 0: i, + ravet0nF = Yo 200k22 T = RC = 200 x 10° x 10 x 107° = 2ms; = = 500 Uo = 2450 V7, t>0r docsity.com Problems 7-47 Uo Dig ee ron ie OO es A IP] % = 599,000 = 200,000 ~ 1206 t = ica x 120 x 10° f er de +0 = 16-16 V, > 0 P7.49 [a] The energy delivered to the 200kQ resistor is equal to the energy stored in the equivalent capcitor. From the solution to Problem 7.48 we have ie su? _ 5(t0 x 107*)(24)? = 2.88 J [b] From the solution to Problem 7.48 we know the voltage on the 15 nF capacitor at t = oo is 16 V. THerefore, the voltage across the 30 nF capacitor at t = co is —16 V. It follows that the total energy trapped is 1 Wetsepea' = 3 (30 x 10~®)(—16)? + 5(i5 x 107°)(16)? = 5.76 J [e] w(0) = 330 x 10-9)(242) = 8.64 J Check: Wrapped + Weise = 5.76 + 2.88 = 8.64 = w(0) P 7.50 [a] t>0 i 4kQ 8kQ 1 ae + 4 Loma) ois? on 7 vg 25nF > Orv vy 25nF cm Vo(0~) = vo(0T) = 0V Uo(00) = 40 V 7 = (8x 10°)(25)x 10° =0.2ms 1/7 = 5000 Vo = (40 — 40e75°%) v, t>0 9d dt ig = 25 x 1079(200,000e5°™) = 5e~50 mA [b] ig = 25 x 10~ v1 = 4(5e 50%) + 40 — 402750% = 40 — 2027500 ig = {c] tx (4) = ig + ic = 2 + 4e0 mA 3 docsity.com